How do I find the integral #int1/((w-4)(w+1))dw# ?
1 Answer
#=1/5ln((w-4)/(w+1))+c# , where#c# is a constant
Explanation :
This type of question usually solve by using Partial Fractions,
#1/((w-4)(w+1))# , it can be written as
#1/((w-4)(w+1))=A/(w-4)+B/(w+1)#
multiplying by
#1=A(w+1)+B(w-4)#
#1=(A+B)w+(A-4B)#
Now comparing coefficient of
#A+B=0# #=># #A=-B# ...........#(i)#
#A-4B=1# ..............#(ii)#
Substituting value of
#-5B=1# #=># #B=-1/5#
from
Now,
#1/((w-4)(w+1))=1/(5(w-4))-1/(5(w+1))#
Integrating both side with respect to
#int1/((w-4)(w+1))dw=int1/(5(w-4))dw-int1/(5(w+1))dw#
#=1/5(ln(w-4)-ln(w+1))+c# , where#c# is a constant
#=1/5ln((w-4)/(w+1))+c# , where#c# is a constant